用RSTAN拟合泊松HMM JAGS模型

问题描述

Walter Zucchini在他的书中 Hidden Markov时间序列模型第129页第8章中的使用R 简介"使用R2OpenBUGS调整Poisson HMM，然后显示代码.我对使用rstan调整相同的模型感兴趣，但是由于我是新使用此软件包的人，因此我对语法没有任何建议.

Walter Zucchini in his book Hidden Markov Models for Time Series An Introduction Using R, in chapter 8 page 129, adjusts a Poisson HMM using R2OpenBUGS, then I show the code. I am interested in adjusting this same model but with rstan, but since I am new using this package, I am not clear about the syntax any suggestion.

数据

dat <- read.table("http://www.hmms-for-time-series.de/second/data/earthquakes.txt")

RJAGS

library(R2jags)
library(rjags)

HMM <- function(){
  for(i in 1:m){
    delta[i] <- 1/m
    v[i] <- 1}
  s[1] ~ dcat(delta[])
  for(i in 2:100){
    s[i] ~ dcat(Gamma[s[i-1],])}
  states[1] ~ dcat(Gamma[s[100],])
  x[1]~dpois(lambda[states[1]])
  for(i in 2:n){
    states[i]~dcat(Gamma[states[i-1],])
    x[i]~dpois(lambda[states[i]])}
  for(i in 1:m){
    tau[i]~dgamma(1,0.08)
    Gamma[i,1:m]~ddirch(v[])}
  lambda[1]<-tau[1]
  for(i in 2:m){
    lambda[i]<-lambda[i-1]+tau[i]}}

x = dat[,2]
n = dim(dat)[1]
m = 2

mod = jags(data = list("x", "n", "m" ), inits = NULL, parameters.to.save = c("lambda","Gamma"),
           model.file = HMM, n.iter = 10000, n.chains = 1)

输出

mod
Inference for Bugs model at "C:/Users/USER/AppData/Local/Temp/RtmpOkrM6m/model36c8429c5442.txt", fit using jags,
 1 chains, each with 10000 iterations (first 5000 discarded), n.thin = 5
 n.sims = 1000 iterations saved
           mu.vect sd.vect    2.5%     25%     50%     75%   97.5%
Gamma[1,1]   0.908   0.044   0.805   0.884   0.915   0.940   0.971
Gamma[2,1]   0.155   0.071   0.045   0.105   0.144   0.195   0.325
Gamma[1,2]   0.092   0.044   0.029   0.060   0.085   0.116   0.195
Gamma[2,2]   0.845   0.071   0.675   0.805   0.856   0.895   0.955
lambda[1]   15.367   0.763  13.766  14.877  15.400  15.894  16.752
lambda[2]   26.001   1.321  23.418  25.171  25.956  26.843  28.717
deviance   645.351   8.697 630.338 639.359 644.512 650.598 665.405

DIC info (using the rule, pD = var(deviance)/2)
pD = 37.8 and DIC = 683.2
DIC is an estimate of expected predictive error (lower deviance is better).

RSTAN

library("rstan")
rstan_options(auto_write = TRUE)
options(mc.cores = parallel::detectCores())

HMM <- '
data{
  int<lower=0> n;    // number of observations (length)
  int<lower=0> x[n]; // observations
  int<lower=1> m;    // number of hidden states
 }

parameters{
  simplex[m] Gamma[n]; // t.p.m
  vector[m] lambda;  // mean of poisson ordered
 }

model{
  vector[m] delta[m];
  vector[m] v[m];
  vector[100] s[100];
  vector[n] states[n];
  vector[m] tau;

  for(i in 1:m){
    delta[i] = 1/m;
    v[i] = 1;}
  s[1] ~ categorical(delta[]);

  for(i in 2:100){
    s[i] ~ categorical(Gamma[s[i-1],]);}
  states[1] ~ categorical(Gamma[s[100],]);
  x[1] ~ poisson(lambda[states[1]]);

  for(i in 2:n){
    states[i] ~ categorical(Gamma[states[i-1],]);
  x[i] ~ poisson(lambda[states[i]])};

  for(i in 1:m){
    tau[i] ~ gamma(1,0.08);
    Gamma[i,1:m] ~ dirichlet(v[]);}
  lambda[1] = tau[1];

  for(i in 2:m){
    lambda[i] = lambda[i-1] + tau[i]};}'

data <- list(n = dim(dat)[1], x = dat[,2], m = 2)
system.time(mod2 <- stan(model_code = HMM, data = data, chains = 1, iter = 1000, thin = 4))
mod2

但是，运行stan模型时会发生错误.

however, an error occurs when running the stan model.

推荐答案

对于从属状态的均值矢量，使用正向算法并以伽马分布作为先验，对simplex[m]对象施加限制，以便概率转移矩阵，其中行之和等于1.获得以下估计.

Using the forward algorithm, and as priors the gamma distribution, for the means vector of the dependent states, and imposing the restriction on the simplex[m] object, for the probability transition matrix, in which the sum by rows equals 1 The following estimates are obtained.

dat <- read.table("http://www.hmms-for-time-series.de/second/data/earthquakes.txt")
stan.data <- list(n=dim(dat)[1], m=2, x=dat$V2)

PHMM <- '
data {
  int<lower=0> n; // length of the time series
  int<lower=0> x[n]; // data
  int<lower=1> m; // number of states
}

parameters{
  simplex[m] Gamma[m]; // tpm
  positive_ordered[m] lambda; // mean of poisson - ordered
}

model{
  vector[m] log_Gamma_tr[m]; // log, transposed tpm
  vector[m] lp; // for forward variables
  vector[m] lp_p1; // for forward variables

  lambda ~ gamma(0.1, 0.01); // assigning exchangeable priors
  //(lambdas´s are ordered for sampling purposes)

  // transposing tpm and taking the log of each entry
  for(i in 1:m)
    for(j in 1:m)
      log_Gamma_tr[j, i] = log(Gamma[i, j]);

  lp = rep_vector(-log(m), m); //

    for(i in 1:n) {
      for(j in 1:m)
        lp_p1[j] = log_sum_exp(log_Gamma_tr[j] + lp) + poisson_lpmf(x[i] | lambda[j]);

      lp = lp_p1;
    }

  target += log_sum_exp(lp);
}'

model <- stan(model_code = PHMM, data = stan.data, iter = 1000, chains = 1)
print(model,digits_summary = 3)

输出

Inference for Stan model: 11fa5b74e5bea2ca840fe5068cb01b7b.
1 chains, each with iter=1000; warmup=500; thin=1;
post-warmup draws per chain=500, total post-warmup draws=500.

               mean se_mean    sd     2.5%      25%      50%      75%    97.5% n_eff  Rhat
Gamma[1,1]    0.907   0.002 0.047    0.797    0.882    0.913    0.941    0.972   387 0.998
Gamma[1,2]    0.093   0.002 0.047    0.028    0.059    0.087    0.118    0.203   387 0.998
Gamma[2,1]    0.147   0.004 0.077    0.041    0.090    0.128    0.190    0.338   447 0.999
Gamma[2,2]    0.853   0.004 0.077    0.662    0.810    0.872    0.910    0.959   447 0.999
lambda[1]    15.159   0.044 0.894   13.208   14.570   15.248   15.791   16.768   407 1.005
lambda[2]    25.770   0.083 1.604   22.900   24.581   25.768   26.838   28.940   371 0.998
lp__       -350.267   0.097 1.463 -353.856 -351.091 -349.948 -349.155 -348.235   230 1.001

Samples were drawn using NUTS(diag_e) at Wed Jan 16 00:35:06 2019.
For each parameter, n_eff is a crude measure of effective sample size,
and Rhat is the potential scale reduction factor on split chains (at
convergence, Rhat=1).

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